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Does a diode in the Vcc line disable clamp diodes to Vcc

Discussion in 'Electronic Design' started by Bill, Sep 15, 2012.

  1. Bill

    Bill Guest

    I have a case where I want to put about 20 microamps into an
    AVR at 12v. AVRs have clamp diodes to both ground and Vcc on
    their digital inputs. It seems to me that a diode inline with
    the Vcc supply would disable the internal clamp diodes of the
    chip. Am I correct on this or is the internal circuitry
    fancier than a simple clamp diode.

    Thanks,
    Bill
     
  2. Jamie

    Jamie Guest

    The input diodes you speak of are connected to the VSS and VDD
    internally and are there for a couple of reasons.

    1. is to prevent over drive of either low or high of course.

    2. Since the internals are CMOS, most likely, it's not a good idea to
    allow inputs to exceed the +/- rails, otherwise, you'll get a latch up
    and a misbehaved chip if input signals for even a moment, exceeding a
    diode drop above the rail voltage.

    You need not to worry about the supply diode you have in line to the
    (+) supply of the chip. It's all about making sure the inputs do not
    exceed the +/- of the rails at the chip. The voltage at the (+) rail of
    the chip, for example, will cause current to flow in the upper diode if
    it reaches a diode drop (~ 0.6) above the (+) rail, at which point the
    input signal will flow through the diode to the (+) rail instead,
    against the supply. The clamping force will still be provided.

    It's possible they're using sk diodes so it could even be closer to .3
    volts when (If) is started.

    You really want to avoid hitting the inputs hard enough to activate
    these diodes. It's not good practice and it will most likely extend the
    skew. If you need to clip an input signal to compress it to a square
    wave for an adequate trigger, do that before with an external circuit.

    Jamie
     
  3. It's perhaps a bit fancier, but yes this will likely "disable" the
    clamp diode to Vdd.

    Meaning that if you pull an input up to 12V, you might think that only
    leakage current would flow back into the Vdd supply. Unfortunately,
    the diode is still in there and will conduct the 12V (less a diode
    drop) to the Vdd of the internals of the chip, and it will probably
    die a painful death.

    As someone else said, this will likely cause latch-up, the Vdd supply
    at the chip Vdd pin will come crashing down to a volt or so as the
    parasitic SCR turns on, and then your series diode will conduct in the
    forward direction (a whole bunch of current).



    Best regards,
    Spehro Pefhany
     
  4. Jamie

    Jamie Guest

    I fail to see how that would happen ?

    If the Vcc(Vdd) is getting supplied via this external diode from the
    supply, the chip is only going to see that VDD there. It will still
    act as the limiting factor for the upper side of the internal clamp
    diode. If anything, if the input signal is getting a reference from the
    supply that maybe even .7 higher, it'll just cause the diode to clamp
    sooner..

    Even if the input some how manages to raise the Vdd, due to internal
    current passing through the clamp to the Vdd, it'll just elevate the
    Vdd. The most that can happen is a slight elevation of Vdd but since
    latch up involves input being above Vdd, I still fail to see how it can
    happen?

    I may not express myself with 101 "Harvard English", but I think I
    get my point across, most of the time.

    Maybe I am blind, due to all the magic smoke that has escaped me over
    the years.

    "Once I thought I was wrong, then I found that every one else was mistaken"

    Jamie
     
  5. fOn Fri, 14 Sep 2012 22:25:08 -0400, the renowned Jamie
    Vdd o +5
    |
    V External diode
    -
    | +19.3
    Vdd pin of chip o-------------------+---------------------+
    | | |
    | | ---
    - | --- bypass
    ^ ||-+ |
    +20V | ||-> ===
    o--------------+--------------|-||-+ GND
    | | |
    | | +---------o
    - | |
    ^ | ||-+
    | | ||<-
    | |-||-+
    | |
    === ===
    GND GND

    The external diode isolates the Vdd pin of the chip from the external
    supply.


    Why slight? A micro, depending on mode, might be drawing only leakage
    current, so the voltage would rise until something breaks down.

    With such a small current (assuming it _is_ limited- it's not clear
    whether the current is limited or the OP wishes to draw a maximum
    current of 20uA or maybe there is another meaning altogether), it
    might not damage the chip, in fact, but it certainly won't allow the
    input to go much above the external supply safely since the internal
    Vdd of the chip sees the input voltage less a diode drop.

    An easy way to do this, as I think JL mentioned, is to use a CD4049/50
    buffer, which has an abs max input voltage of 18V.

    But if the 12V signal is coming from the outside world and 20uA is
    allowable current draw, there are lots of ways of handling it more
    safely such as dividing it down or using a BJT.



    Best regards,
    Spehro Pefhany
     
  6. Phil Allison

    Phil Allison Guest

    "Tim Wescott"

    ** It does, if you read it properly.


    ** The OP is talking about the "digital inputs ".


    ** But you remove the current sinking effect of PSU source impedance from
    the Vcc pin.



    .... Phil
     
  7. Jamie

    Jamie Guest

    Hmm, it seems that our descriptions of what is and what isn't, is
    getting mixed in translation.

    If I read you correctly, you now agree with me, at first it seem to be
    the opposite?

    Oh well, no matter anyway. I think most understand the logic.

    On to the next...


    Jamie
     
  8. Jasen Betts

    Jasen Betts Guest

    All inputs except for the reset pin.
    No. it's that simple,

    The diode will continue to function. depending on the model of AVR
    and what it's doing 20 microamps may be more than the current the
    AVR is consuming in which case you'll loose supply voltage regulation
    for the chip.
     
  9. TTman

    TTman Guest

    No problem.... the internal diodes will take 20uA easy.

    Is this for a battery voltage detect system ?
     
  10. I have not tested the AVR, but other chips I have tested don't behave
    quite like a small diode. There is quite a lot of ohmic resistance in
    there IIRC, at least on the "diode" to Vdd. According to what IIRC JT
    said, the exact ESD-protection circuitry is generally proprietary.

    That said, usually you can think of it as a pair of diodes, with the
    caveat that running current through those diodes can make strange
    things happen, from slight errors in an ADC to outright destruction of
    the chip at very high peak current. Another slightly more accurate
    (IME) model is a diode to Vdd with a resistor in series, and (more or
    less) a diode to Vss.


    Best regards,
    Spehro Pefhany
     
  11. Bill

    Bill Guest

    Thanks for the answers.

    I had modeled this in LTSPICE and had found that
    the clamp didn't protect it if there was a diode
    in the Vcc line.

    I blew it. I had included the impedance of the signal
    source but I forgot to include the internal load of the
    circuit of the circuit in the IC. I have now put a
    500 ohm resistor in to represent this load and the
    diode clamps are doing their job in the LTSPICE model.

    Good Luck,
    Bill
     
  12. Jamie

    Jamie Guest

    They call that a oops. Normally if you have a bypass cap there
    and only present a transient, it should model ok. Like I said, should.

    I don't use Ltspice or spice in general all that much, I've done
    everything with a calculator, paper, bread boards and test jigs. It
    wasn't up to about 2 or so years ago I even tried Ltspice.

    Years ago I tried using emulators, but the computers were so slow and
    software so bad that I was able to get my answer faster by hand.

    WHen my little one when to trade school, is when I saw the use of
    workbench, I really didn't think much of that. The graphical view
    of the circuit was more like an arcade game, didn't really turn me on.

    Now days, I walk around at work with my TI in my shirt pocket and a
    small laptop that is easy to carry. lately the bench hasn't gotten much
    exercise.

    Jamie
     
  13. Guest

    What do you mean by emulators? Not generally but specifically a product, can't imagine the use for something like that, like the resistor box joke thing.
     
  14. josephkk

    josephkk Guest

    Well not really. !
    1.) use a Schottky diode if you can and a circuit to limit Vdd, like a
    transistor beefed up TL431.
    2.) it is not really a diode but a more complex circuit that acts similar
    to a diode and is very fast (must be, as ESD is very fast).

    ?-)
     
  15. Jamie

    Jamie Guest

    My son used a program while in trade school which was called "Work
    Bench", it actually showed you a bread board, jumper wires, real view of
    a DMM and explosions and smoke as things didn't turn out right.

    Maybe I have the product name incorrect but that is what I remember
    seeing it called on the screen..

    But when it came time to build a AM&FM receiver kit, he presented it to
    me after he had assembled it with a few minor problems. While I helped
    him correct those problems, I also did a little modification on it to
    bump up gain and quiet the receiver.

    He got to enter that in a parents get together for public view and it
    ended up in a glass case at school for a while, along with a couple
    other kids, which also seem to have had some help that were in the business.

    I really don't think the instructor caught on, maybe he did and didn't
    say any however, he may have noticed those transistor types I changed
    if he really looked a littler closer, along with some added by passes
    etc. :)

    Jamie
     
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